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$\bf证明$  由于$m\left( {E\left( {{f_n} \nrightarrow f} \right)} \right) = 0$，则我们不妨设$\left\{ {{f_n}\left( x \right)} \right\}$处处收敛于$f(x)$，此时\[E = \bigcup\limits_{m = 1}^\infty  {\bigcap\limits_{n = m}^\infty  {E\left( {\left| {{f_n} - f} \right| < \frac{1}{k}} \right)} } ,k \in {N_ + }\]记\[{B_{m,k}} = \bigcap\limits_{n = m}^\infty  {{E_{n,k}}}  = E\left( {\left| {{f_n} - f} \right| < \frac{1}{k},n \ge m} \right)\]其中${E_{n,k}} = E\left( {\left| {{f_n} - f} \right| < \frac{1}{k}} \right)$，则对于固定的$k$，${B_{m,k}}$是单调递增的集合列，并且$E = \bigcup\limits_{m = 1}^\infty  {{B_{m,k}}} $，所以我们有$m\left( E \right) = \lim \limits_{m \to \infty } m\left( {{B_{m,k}}} \right)$，而$m\left( E \right) < \infty $，则对任给$\delta  > 0$，存在${n_k}\left( { > {n_{k - 1}}} \right)$，使得\[m\left( E \right) - m\left( {{B_{{n_k},k}}} \right) < \frac{\delta }{{{2^k}}}\]

令\[F = \bigcap\limits_{k = 1}^\infty  {{B_{{n_k},k}}}  = E\left( {\left| {{f_n} - f} \right| < \frac{1}{k},n \ge {n_k}} \right)\]

则我们有\[m\left( {E\backslash F} \right) = m\left( {\bigcup\limits_{k = 1}^\infty  {\left( {E\backslash {B_{{n_k},k}}} \right)} } \right) \le \sum\limits_{k = 1}^\infty  {m\left( {E\backslash {B_{{n_k},k}}} \right)}  < \delta \]

以及对任给的$\varepsilon  > 0$，存在${k_0} > \frac{1}{\varepsilon }$，使得当$n \ge {n_{{k_0}}}$时，对任意的$x \in F$，有\[\left| {{f_n}\left( x \right) - f\left( x \right)} \right| < \frac{1}{{{k_0}}} < \varepsilon \]

所以$\left\{ {{f_n}\left( x \right)} \right\}$在$F$上一致收敛于$f(x)$